All Questions
Tagged with derham-cohomology cohomology
25 questions
18
votes
1
answer
682
views
De Rham via topoi
Étale cohomology of schemes $X$ is constructed as follows: one associates to $X$ the so-called étale topos of $X$, and then one just takes the sheaf cohomology of that topos.
Is it possible to ...
15
votes
1
answer
2k
views
Artin vanishing for Stein manifolds and restriction maps
In the setting of complex Stein manifolds $X$ of complex dimension $d$, the theorem of Andreotti--Frankel implies the vanishing of the singular cohomology group $H^i(X,\mathbb Z)=0$ for $i>d$. With ...
13
votes
1
answer
725
views
Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex?
If $G$ is a discrete or a Lie Group acting smoothly on a manifold $M$, we can define the algebra of $G$-invariant de Rham classes, $H(M)^G$, and we can also consider the cohomology of the sub-complex ...
12
votes
0
answers
301
views
Is there a differential form which corresponds to an eigenvalue of the homomorphism in cohomology?
Let $M$ be a closed manifold and $f:M\to M$ be a diffeomorphism. Suppose the homomorphism $f^*:H^k(M;\mathbb R)\to H^k(M;\mathbb R)$ has an eigenvalue $\lambda\in\mathbb{R}$. Note that $\lambda$ is ...
9
votes
2
answers
1k
views
Hodge dual of de Rham cohomology and singular cohomology
We know that the de Rham cohomology is isomorphic to the singular cohomology, does the Hodge dual of differential forms induce a dual operation on de Rham cohomology, hence also on singular cohomology?...
9
votes
1
answer
392
views
Integrating hypercohomology classes
Let $X$ be a complex variety. By Poincare's lemma, its singular cohomology can be computed as hypercohomology of the holomorphic de Rham complex (viewing $X$ as a complex manifold)
$$\text{H}^\cdot(X,\...
9
votes
0
answers
347
views
Is there a Hodge isomorphism theorem for part-tangential, part-normal, harmonic differential forms?
Let $M$ be an oriented compact Riemannian $n$-manifold with boundary $\partial M$. A differential $p$-form $\omega$ on $M$ is normal if $i^* \omega = 0$ holds, tangential if $i^* \star \omega = 0$ ...
7
votes
1
answer
2k
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What is the scope of validity of Kunneth formula for de Rham?
In books like Bott-Tu or all pdf texts I have found on internet, the Kunneth formula for manifolds $M$ and $N$ and their de Rham cohomology
$$ H^{\bullet}_{dR}(M \times N) \simeq H^{\bullet}_{dR}(M) \...
6
votes
1
answer
2k
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Integration currents vs Poincaré dual
Let $M$ be a complex manifold of dimension $n$ and $S \subset M$ a closed complex submanifold of complex codimension $r$. Let $[S] \in H_{2r}(S)$ be the fundamental class of $S$.
We have the ...
6
votes
1
answer
374
views
De Rham cohomology of Lie groupoid
Let $G$ be a Lie group acting on a manifold $M$.
Consider the transformation groupoid $\mathcal{G}=(G\times M\rightrightarrows M)$. We have the notion of de Rham cohomology of a Lie groupoid by ...
5
votes
1
answer
228
views
Which compact (finite dimensional) Lie groups have $H^1_{DR}(G)\neq 0$
In particular, I am wondering if $H^1_{DR}(G)\neq 0$ implies that the group can written as a semidirect product of $\mathbb{S^1}$ and something else, with the $\mathbb{S^1}$ factor being responsible ...
5
votes
2
answers
2k
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Global Definition of the Dolbeault Complex of a Vector Bundle
For an $2n$-dimensional complex manifold $M$, and a smooth vector bundle $E$ over $M$, it is well-known (see Voisin, Huybrechts) that there exists an operator $\overline{\partial}$, built locally from ...
5
votes
0
answers
304
views
Generalization of de Rham cohomology, or cohomology for non-smooth case
Let $\Omega\subseteq \mathbb{R}^{3}$ be a contractible region and $f:\Omega\to \mathbb{R}^{3}$ be a vector field on $f$. If $f$ is smooth vector field and $\nabla\cdot f=0$, then $f=\nabla\times g$ ...
4
votes
0
answers
173
views
Spencer complex and de Rham Complex
in those lectures notes written by Claude Sabbbah: https://perso.pages.math.cnrs.fr/users/claude.sabbah/livres/sabbah_nankai110705.pdf
there is the proposition 1.4.4 where he says that there is a ...
4
votes
0
answers
109
views
Generalized de Rham cohomology on product bundle giving specified cohomology
Given a compact, smooth manifold $M$ and a real vector bundle $E \to M$ (in general not flat). There already have been numerous questions about how to equip the space $\bigoplus_k \Gamma(\Lambda^k T^* ...
4
votes
0
answers
441
views
Finite Field Varieties and the de Rham Complex of Kähler Differentials
In an answer to a previous question I asked, where the Kahler differentials of a variety over a finite field were discussed, it was stated that:
You can certainly define de Rham cohomology using ...
3
votes
1
answer
329
views
Calculation of Frobenius on de Rham cohomology of elliptic curves with good reduction
I'm reading "An introduction to the theory of $p$-adic representations" by Laurent Berger. In the page 14 it says:
Suppose $F$ is an unramified finite extension of $\mathbb Q_p$ and $E$ is ...
3
votes
1
answer
167
views
Models for computing cohomology of Lie groupoids
Given a Lie groupoid $\mathcal{G}=[\mathcal{G}_1\rightrightarrows \mathcal{G}_0]$, let $\mathcal{G}_\bullet$ be the associated simplicial manifold.
Let $\Omega^\bullet(\mathcal{G}_\bullet)$ be the ...
3
votes
0
answers
217
views
Reference: Relative cohomology of a morphism
Let $f\colon Y \to X$ be a morphism of schemes, the inverse image in $K$-theory always fit into a long exact sequence
$$
\cdots \to K_i(f)\to K_i(X) \xrightarrow {f^*} K_i(Y)\to \cdots
$$
where the ...
2
votes
1
answer
366
views
Easier ways to compute homology/cohomology by adding extra structure
Suppose $X$ is a topological space and I want to talk about its “homology”.
There is this notion of singular homology obtained from singular chain complex. This is not very easy to compute.
Suppose ...
2
votes
0
answers
235
views
Mayer-Vietoris on Fibered Products
Suppose $M \xrightarrow{p} X$ is a surjective submersion of smooth manifolds. Define the fibered product in this case: $$M \times_X M = \{ (m_1, m_2) \in M \times M | p(m_1) = p(m_2) \}$$
and let $U =...
1
vote
0
answers
303
views
Berthelot-Ogus comparison isomorphism
On the link, page, $ 2 $, the Berthlot-Ogus isomorphism theorem is stated as follows,
We have a canonical isomorphism, $$ \rho_{\mathrm{cris}} \ : \ H_{\mathrm{cris}}^{i} (X) \otimes_{K_ {0}} K \to H_{...
1
vote
0
answers
217
views
Find torsion classes using flat bundles
My question refers to a discussion from this older thread on Neron-Severi group of a Kähler manifold. In the comments below Ted Shifrin's answer there arose a discussion when the map $H^2(X,\mathbb{Z}...
1
vote
0
answers
348
views
Vanishing of cohomology groups
Given a smooth hypersurface $X \subset \mathbb{P}^{2p+1}_{\mathbb{C}}$ of degree $d \ge 5$ (with $p\ge 2$), when can we say that
$H^{p+1}(\mathcal{O}_X)=H^{p+1}(\Omega^1_X)=...=H^{p+1}(\Omega^{p-2}...
-2
votes
1
answer
89
views
Alternating property of H_2(T, Z)
Let us consider the torus $T = S^1_X \times S^1_Y$, where the former $S^1_X$ has the coordinate $X$, whereas the latter $S^1_Y$ has $Y$. We have an alternate property $dX \wedge dY = - dY \wedge dX \...